Proposition 10

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h}$ a nilpotent subalgebra of $\mathfrak{g}$. Denote by P the set of maps from $\mathfrak{h}$ to $k$.

(i) For $\lambda , \mu\in P$, $[\mathfrak{g}^{\lambda}(\mathfrak{h}),\mathfrak{g}^\mu(\mathfrak{h})]\subset \mathfrak{g}^{\lambda+\mu}(\mathfrak{h})$; in particular, $\mathfrak{g}^0(\mathfrak{h})$ is a Lie subalgebra of $\mathfrak{g}$ containing $\mathfrak{h}$, and the $\mathfrak{g}^{\lambda}(\mathfrak{h})$ are stable under ad $\mathfrak{g}^0(\mathfrak{h})$. Moreover, $\mathfrak{g}^0(\mathfrak{h})$ is its own normalizer in $\mathfrak{g}$.

(ii) If M is a $\mathfrak{g}$-module, $\mathfrak{g}^{\lambda}(\mathfrak{h})M^\mu(\mathfrak{h})\subset M^{\lambda+\mu}(\mathfrak{h})$ for $\lambda , \mu\in P$; in particular, each $M^{\lambda}(\mathfrak{h})$ is a $\mathfrak{g}^0(\mathfrak{h})$-module.

(iii) If B is a bilinear form on $\mathfrak{g}$ invariant under $\mathfrak{h},\mathfrak{g}^{\lambda}(\mathfrak{h})$ and $\mathfrak{g}^\mu(\mathfrak{h})$ are orthogonal relative to B for $\lambda +\mu\not= 0$. Assume that B is non-degenerate. Then, for all $\lambda \in P$, the restriction of B to $\mathfrak{g}^{\lambda}(\mathfrak{h})\times \mathfrak{g}^{-\lambda}(\mathfrak{h})$ is non-degenerate; in particular, the restriction of B to $\mathfrak{g}^0(\mathfrak{h})\times \mathfrak{g}^0(\mathfrak{h})$ is non-degenerate.

(iv) Assume that $k$ is of characteristic 0. Then, if $x\in \mathfrak{g}^{\lambda}(\mathfrak{h})$ with $\lambda \not= 0$, ad $x$ is nilpotent.

The map $(x, y) \rightarrow [x, y]$ from $\mathfrak{g}\times \mathfrak{g}$ to $\mathfrak{g}$ is $\mathfrak{g}$-invariant by the Jacobi identity, hence $\mathfrak{h}$-invariant. The first part of (i) thus follows from Prop. 2 (ii). Part (ii) is proved similarly.

If $x$ belongs to the normalizer of $\mathfrak{g}^0(\mathfrak{h})$ in $\mathfrak{g}$, (ad $y$)$.x=-[x, y]\in \mathfrak{g}^0(\mathfrak{h})$ for all $y\in \mathfrak{h}$, so (ad $y$)$^n.x= 0$ for $n$ sufficiently large. This proves that $x\in \mathfrak{g}^0(\mathfrak{h})$. Assertion (i) is now completely proved.

Assertion (iii) follows from Prop. 9 (v).

To prove (iv), we can assume that $k$ is algebraically closed. Let $x\in \mathfrak{g}^{\lambda}(\mathfrak{h})$, with $\lambda \not= 0$. For all $\mu\in P$ and any integer $n\geq 0$, (ad $x$)$^n\mathfrak{g}^\mu(\mathfrak{h})\subset \mathfrak{g}^{\mu+n\lambda}(\mathfrak{h})$; let $P_1$ be the finite set of $\mu\in P$ such that $\mathfrak{g}^\mu(\mathfrak{h})\not= 0$; if $k$ is of characteristic 0 and $\lambda \not= 0, (P_1+n\lambda )\cap P_1=\emptyset$ for $n$ sufficiently large, so (ad $x$)$^n= 0$.